Toss a Coin Six Times

Date: 02/07/98 at 16:59:43
From: Ruth Beldon
Subject: Coin tossing probabilities
A. Suppose a coin is tossed 6 times. What is the probability that 6
heads will occur? (Answer: 1/64)
B. What is the probablity that 3 heads will occur? (Book answer: 5/16)
6/3 x 1/2 to 3rd power x 1/2 to 3rd power = 20x1/8x1/8 = 5/16
C. X = 2 6/2x 1/2 squared x 1/2 to 4th = 15x1/4x1/16 = 15/64
My question is: where did the 20 come from in part B and the 15 in
part C? How was this answer arrived at?
Thank you,
R. Beldon

Date: 02/07/98 at 18:29:05
From: Doctor Mitteldorf
Subject: Re: Coin tossing probabilities
Dear Ruth,
The way you calculate probabilities for n coin tosses is to count the
different ways (different combinations) that the event you're looking
at could happen.
Say there are 6 tosses. The first toss can be either heads or tails.
The second can be either heads or tails. 2*2 = 4. The third can be
either heads or tails... so you end up with 2^6 = 64 possibilities.
Only one of these has all heads. But there are more ways that you
could get 3 heads. It could be the first, second, and third, or the
first, second and fourth that are heads. Or maybe the first, second
and fifth.
Here's a complete list:
123,124,125,126,134,135,136,145,146,156,234,235,
236,245,246,256,345,346,356,456.
That's 20 possibilities out of 64, or 20/64 = 5/16.
The answer is related to Pascal's triangle. The 6th row is
1 6 15 20 15 6 1
The numbers add up to 64, and the middle one is 20. There is a formula
for these numbers, which your book is referring to:
rth number in nth row of Pascal Triangle (counting from zero):
n!
---------------
(n-r)! r!
In your case, 6*5*4*3*2*1 in the numerator, 3*2*1 and 3*2*1 again in
the denominator.
-Doctor Mitteldorf, The Math Forum
Check out our web site! http://mathforum.org/dr.math/